Given an array arr[], the task is to count all the pairs whose xor gives the unique prime, i.e. no two pairs should give the same prime.
Examples:
Input: arr[] = {2, 3, 4, 5, 6, 7, 8, 9}
Output: 6
(2, 5), (2, 7), (2, 9), (4, 6), (4, 7) and (4, 9) are the only pairs whose XORs are primes i.e. 7, 5, 11, 2, 3 and 13 respectively.
Input: arr[] = {10, 12, 23, 45, 5, 6}
Output: 4
Approach: Iterating every possible pair and check whether the xor of the current pair is a prime. If its a prime then update count = count + 1 and also save the prime in an unordered_map in order to keep track of the repeating primes. Print the count in the end.
Below is the implementation of the above approach:
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;
// Function that returns true if n is prime bool isPrime( int n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to return the count of valid pairs int countPairs( int a[], int n)
{ int count = 0;
unordered_map< int , int > m;
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == 0) {
m[(a[i] ^ a[j])]++;
count++;
}
}
}
return count;
} // Driver code int main()
{ int a[] = { 10, 12, 23, 45, 5, 6 };
int n = sizeof (a) / sizeof (a[0]);
cout << countPairs(a, n);
} |
// Java implementation of above approach import java.util.*;
class solution
{ // Function that returns true if n is prime static boolean isPrime( int n)
{ // Corner cases
if (n <= 1 )
return false ;
if (n <= 3 )
return true ;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0 )
return false ;
for ( int i = 5 ; i * i <= n; i = i + 6 )
if (n % i == 0 || n % (i + 2 ) == 0 )
return false ;
return true ;
} // Function to return the count of valid pairs static int countPairs( int a[], int n)
{ int count = 0 ;
Map<Integer, Integer> m= new HashMap< Integer,Integer>();
for ( int i = 0 ; i < n - 1 ; i++) {
for ( int j = i + 1 ; j < n; j++) {
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && m.get(a[i] ^ a[j]) == null ) {
m.put((a[i] ^ a[j]), 1 );
count++;
}
}
}
return count;
} // Driver code public static void main(String args[])
{ int a[] = { 10 , 12 , 23 , 45 , 5 , 6 };
int n = a.length;
System.out.println(countPairs(a, n));
} } // This code is contributed by Arnab Kundu |
# Python3 implementation of above approach # Function that returns true if n is prime def isPrime(n):
# Corner cases
if n < = 1 :
return False
if n < = 3 :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if n % 2 = = 0 or n % 3 = = 0 :
return False
i = 5
while i * i < = n:
if n % i = = 0 or n % (i + 2 ) = = 0 :
return False
i + = 6
return True
# Function to return the count of valid pairs def countPairs(a, n):
count = 0
m = dict ()
for i in range (n - 1 ):
for j in range (i + 1 , n):
# If xor(a[i], a[j]) is prime and unique
if isPrime(a[i] ^ a[j]) and m.get(a[i] ^ a[j], 0 ) = = 0 :
m[(a[i] ^ a[j])] = 1
count + = 1
return count
# Driver code a = [ 10 , 12 , 23 , 45 , 5 , 6 ]
n = len (a)
print (countPairs(a, n))
# This code is contributed by # Rajnis09 |
// C# implementation of the approach using System ;
using System.Collections ;
class solution
{ // Function that returns true if n is prime
static bool isPrime( int n)
{
// Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
}
// Function to return the count of valid pairs
static int countPairs( int []a, int n)
{
int count = 0;
Hashtable m= new Hashtable();
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == null )
{
m.Add((a[i] ^ a[j]),1);
count++;
}
}
}
return count;
}
// Driver code
public static void Main()
{
int []a = { 10, 12, 23, 45, 5, 6 };
int n = a.Length;
Console.WriteLine(countPairs(a, n));
}
// This code is contributed by Ryuga
} |
<script> // Javascript implementation of above approach // Function that returns true if n is prime
function isPrime(n)
{
// Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
}
// Function to return the count of valid pairs
function countPairs(a,n)
{
let count = 0;
let m = new Map();
for (let i = 0; i < n - 1; i++)
{
for (let j = i + 1; j < n; j++)
{
// If xor(a[i], a[j]) is prime and unique
if (isPrime(a[i] ^ a[j]) && !m.has(a[i] ^ a[j]))
{
m.set((a[i] ^ a[j]), 1);
count++;
}
}
}
return count;
}
// Driver code
let a = [10, 12, 23, 45, 5, 6];
let n = a.length;
document.write(countPairs(a, n));
// This code is contributed by unknown2108
</script> |
4
Time Complexity: O(n5/2)
Auxiliary Space: O(n2)